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Dmitriy789 [7]
3 years ago
12

Which of these choices show a pair of equivalent expressions? CHECK ALL THAT APPLY

Mathematics
1 answer:
aleksklad [387]3 years ago
6 0

Answer:

Option C & D

Step-by-step explanation:

<u>__________________________________________________________</u>

<u>FACTS TO KNOW BEFORE SOLVING</u> :-

Lets say there's an expression x^{\frac{a}{b} } .

It can be also written it as → (\sqrt[b]{x} )^{a} and vice-versa.

<u>__________________________________________________________</u>

Lets come to the question.

  • In option A , 7^{\frac{5}{7} } can be also written as (\sqrt[7]{7} )^{5} . But it doesn't match with the other expression given. Hence they are not equivalent expressions.
  • In option B , 6^{\frac{2}{7} } can be also written as (\sqrt[7]{6} )^{2} . But it doesn't match with the other expression given. Hence they are not equivalent expressions.
  • In option C , 5^{\frac{3}{2} } can be also written as (\sqrt{5} )^{3}. Hence they are equivalent expressions.
  • In option D , (\sqrt[4]{81} )^{5} can be also written as 81^{\frac{5}{4} } . Hence they are equivalent expressions

Thus , Option C & D are correct.

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The commuting time for a student to travel from home to a college campus is normally distributed with a mean of 30 minutes and a
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Answer:

0.1587

Step-by-step explanation:

Let X be the commuting time for the student. We know that X\sim N(30, (5)^2). Then, the normal probability density function for the random variable X is given by

f(x) = \frac{1}{\sqrt{2\pi(5)^{2}}}\exp[-\frac{(x-\mu)^{2}}{2(5)^{2}}]. We are seeking the probability P(X>35) because the student leaves home at 8:25 A.M., we want to know the probability that the student will arrive at the college campus later than 9 A.M. and between 8:25 A.M. and 9 A.M. there are 35 minutes of difference. So,

 

P(X>35) = \int\limits_{35}^{\infty}f(x)dx = 0.1587

To find this probability you can use either a table from a book or a programming language. We have used the R statistical programming language an the instruction pnorm(35, mean = 30, sd = 5, lower.tail = F)

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4 years ago
A car is designed to last an average of 12 years with a standard deviation of 0.8 years. What is the probability that a car will
Scorpion4ik [409]
Due to the common probability formula I can solve that task. Refering to the task you gave, we have all the information we need. We solve it like that: \frac{10-12}{0.8} = -2.5; Probability(-2.5)=.0.00621; Probability = 0.621%
So the answer is the first option <span>0.621%</span>
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where Robert's spending is greater than Rosa's spending.

Allowance of Rosa = Allowance of Robert

let x = allowance

Rosa : video games = 0.4(x) ; pizza = 2/5 (x)
Robert : video games = 1/2 (x) ; pizza = 0.25 (x)

let us assume that their allowance is 100 each week. so, x = 100

Rosa : video games = 0.40(100) = 40
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total spending: 40 + 40 = 80

Robert : video games = 1/2 (100) = 50
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Spending on video games
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Robert spent more of his allowance on video games than Rosa.

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